Question

Phân tích đặt nhân tử chung:

a) a³ (c - b) + b³ (a - c) + c³ (b - a)

b) xy ( x - y) + yz (y - z) + zx (z - x)

c) x ( y² - z²) + y ( z² - x²) + z ( x - y)

HELLP ME T^T

Answer 1

Em thử, sai thì thôi

a) Đặt c - b =x; a - c = y suy ra b - a = -(x+y)

Ta có \(a^3x+b^3y-c^3\left(x+y\right)\)

\(=x\left(a-c\right)\left(a^2+ac+c^2\right)+y\left(b-c\right)\left(b^2+bc+c^2\right)\)

\(=\left(c-b\right)\left(a-c\right)\left(a^2+ac+c^2\right)-\left(a-c\right)\left(c-b\right)\left(b^2+bc+c^2\right)\)

\(=\left(a-c\right)\left(c-b\right)\left(a^2+ac-b^2-bc\right)\)

\(=\left(a-b\right)\left(a-c\right)\left(c-b\right)\left(a+b+c\right)\)

Answer 2

b) tương tự cũng phải đặt:v

x - y = a; y - z = b thì: z - x = -(a+b)

\(xya+yzb-zx\left(a+b\right)=xya-xza+yzb-xzb\)

\(=xa\left(y-z\right)+zb\left(y-x\right)\)

\(=x\left(x-y\right)\left(y-z\right)-z\left(y-z\right)\left(x-y\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)